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Recursion Problem | AMC 10A, 2019| Problem No 15

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Try this beautiful Problem on Algebra based on Recursion from AMC 10 A, 2019. You may use sequential hints to solve the problem.

Recursion- AMC-10A, 2019- Problem 15

A sequence of numbers is defined recursively by $a_{1}=1, a_{2}=\frac{3}{7},$ and $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

  • $2020$
  • $4039$
  • $6057$
  • $6061$
  • $8078$

Key Concepts


Recursive formula

Suggested Book | Source | Answer

Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2019 Problem-15

Check the answer here, but try the problem first


Try with Hints

First Hint

The given expression is $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$ and given that $a_{1}=1, a_{2}=\frac{3}{7}$. we have to find out \(a_{2019}\)?

at first we may use recursive formula we can find out \(a_3\) , \(a_4\) with the help of \(a_1\), \(a_2\). later we can find out \(a_n\)

Now can you finish the problem?

Second Hint

Given that $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

Now \(n=3\) then $a_{3}=\frac{a_{(3-2)} \cdot a_{(3-1)}}{2 a_{(3-2)}-a_{(3-1)}}$

\(\Rightarrow\) $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

\(\Rightarrow\) $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

\(\Rightarrow\) $a_{3}=\frac{1*\frac{3}{7}}{2*1-\frac{3}{7}}$

\(\Rightarrow\) $a_{3}=\frac{3}{7}$

Similarly if we put \(n=4\) we get \(a_4=\frac{3}{15}\) (where $a_{1}=1, a_{2}=\frac{3}{7}$,\(a_3=\frac{3}{7}\))

Continue this way we $a_{n}=\frac{3}{4 n-1}$

So can you find out the value of \(a_{2019}\)?

Now Can you finish the Problem?

Third Hint

Now $a_{n}=\frac{3}{4 n-1}$

Put \(n=2019\)

$a_{2019}=\frac{3}{8075}$ which is the form of \(\frac{p}{q}\)

Therefore \(p+q=8078\)

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